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**tom-kelleher**on Monty Hall in the Wild · 2018-06-08T20:29:12.526Z · score: 1 (1 votes) · LW · GW

I saw a variation of this explanation that I liked, and made it more intuitive. In helps if we jump from 3 choices to 100.

And as we proceed, think of making the first choice of door A as splitting the group into two sets: Picked and Not-Picked. The "Picked" group always contains one door, the Not-Picked group contains the rest (two doors for the usual version, and 99 in my 100-door version).

And rather than say that Monty Hall "opens *one* of the doors" from the Not-Picked set, let's say he's "opening *all the bogus* doors" from the Not-Picked set. Right? He always opens the door with nothing in it, leaving one remaining door. There's only two in the Not-Picked set, but he always opens one that's bogus.

With that setup, you make your choice. You pick Door #1 out of the 100 doors.

You probably picked wrong; that's no surprise. There's a 99% likelihood that the prize is somewhere in the Not-Picked set of 99 doors. We don't know which one, but hang tight.

Now Monty *opens all the bogus doors* in the Not-Picked set...leaving just one unopened.

That 99% probability of being in the Not-Picked set is *still true--*but now it's all collapsed into that one remaining closed door. That's the only remaining place it could be hiding. Your door choice didn't become any "luckier" just because Monty opened bogus doors.

So the smart money says, switch doors.